# Find the center and foci of ? 9x²+4y²-36x+24y+36=0

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Feb 9, 2018

Center is$\left(2 , - 3\right)$ and focii are $\left(2 , - 3 + \sqrt{5}\right)$ and $\left(2 , - 3 - \sqrt{5}\right)$.

#### Explanation:

Let us write $9 {x}^{2} + 4 {y}^{2} - 36 x + 24 y + 36 = 0$ in general form. In general form equation is ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$, where $\left(h , k\right)$ is center of ellipse and larger of $a$ and $b$ is half the major axis and smaller one is half the minor axis.

The equation can be written as

$9 \left({x}^{2} - 4 x\right) + 4 \left({y}^{2} + 6 y\right) = - 36$

or $9 \left({x}^{2} - 4 x + 4\right) + 4 \left({y}^{2} + 6 y + 9\right) = - 36 + 36 + 36$

or 9(x-2))^2+4(y+3)^2=36

or ${\left(x - 2\right)}^{2} / {2}^{2} + {\left(y + 3\right)}^{2} / {3}^{2} = 1$

Hence center of ellipse is $\left(2 , - 3\right)$, major axis is $6$ parallel to $y$-axis and minor axis is $4$ parallel to $x$-axis.

Hence, we get eccentricity $e = \sqrt{1 - {4}^{2} / {6}^{2}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$

and focii are at a distance of $b e$ i.e. $3 \times \frac{\sqrt{5}}{3}$ i.e. $\sqrt{5}$ on either side of center parallel to major axis i.e.

focii are $\left(2 , - 3 + \sqrt{5}\right)$ and $\left(2 , - 3 - \sqrt{5}\right)$.

graph{(9x^2+4y^2-36x+24y+36)((x-2)^2+(y+3)^2-0.03)((x-2)^2+(y+3+sqrt5)^2-0.03)((x-2)^2+(y+3-sqrt5)^2-0.03)=0 [-6, 10, -7, 1]}

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